Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $p = \dfrac{-t^2 - 4t}{5t - 15} \div \dfrac{t^3 + 14t^2 + 40t}{t^2 - 11t + 24} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{-t^2 - 4t}{5t - 15} \times \dfrac{t^2 - 11t + 24}{t^3 + 14t^2 + 40t} $ First factor out any common factors. $p = \dfrac{-t(t + 4)}{5(t - 3)} \times \dfrac{t^2 - 11t + 24}{t(t^2 + 14t + 40)} $ Then factor the quadratic expressions. $p = \dfrac {-t(t + 4)} {5(t - 3)} \times \dfrac {(t - 3)(t - 8)} {t(t + 4)(t + 10)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac {-t(t + 4) \times (t - 3)(t - 8) } {5(t - 3) \times t(t + 4)(t + 10) } $ $p = \dfrac {-t(t - 3)(t - 8)(t + 4)} {5t(t + 4)(t + 10)(t - 3)} $ Notice that $(t + 4)$ and $(t - 3)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac {-t(t - 3)(t - 8)\cancel{(t + 4)}} {5t\cancel{(t + 4)}(t + 10)(t - 3)} $ We are dividing by $t + 4$ , so $t + 4 \neq 0$ Therefore, $t \neq -4$ $p = \dfrac {-t\cancel{(t - 3)}(t - 8)\cancel{(t + 4)}} {5t\cancel{(t + 4)}(t + 10)\cancel{(t - 3)}} $ We are dividing by $t - 3$ , so $t - 3 \neq 0$ Therefore, $t \neq 3$ $p = \dfrac {-t(t - 8)} {5t(t + 10)} $ $ p = \dfrac{-(t - 8)}{5(t + 10)}; t \neq -4; t \neq 3 $